I have three three questions, the first two of which probably have the same answer and the third of which is more vague.

For a set $A$ let $L_\alpha(A)$ be the constructible universe up to $\alpha$, built from $A$ as a set (and not a predicate). Further let $X = (B, f)$ where $B$ is a transitive set and $f$ is a bijection from $\omega$ to $B$.

Also assume that the background universe has whatever large cardinals you would like (or that would be helpful). In particular though there is at least one inaccessible cardinal in $L$.

(1) Suppose $L_\alpha\models ZFC$. Is it the case that $\omega_1^L = \omega_1^{L_\alpha}$?

(2) Suppose $L_\alpha(X)\models ZFC$. Is it the case that $\omega_1^{L(X)} = \omega_1^{L_\alpha(X)}$?

(3) If the answer to (1), (2) is yes, is there any simpler way for $L_\alpha$ to know that $\omega_1^{L_\alpha} = \omega_1^L$ (other than $L_\alpha\models ZFC$)?

Finally I will just make one observation to highlight why this question isn't trivial. If you replace $ZFC$ with $KP$ then there are many countable admissible sets $L_\alpha\models KP$ with countable (in $L$) ordinals $\beta\in L_\alpha$ such that $L_\alpha \models \omega_1 = \beta$.

2 Answers
2

The answer is no. If there is a transitive set model $M$ of set theory (and this is all you need), then if $\alpha$ is its height (that is, if $\alpha=\mathsf{ORD}^M$), then $L_\alpha$ is a model of $\mathsf{ZFC}+V=L$. Note that the assumption is strictly stronger than the existence of an $\omega$-model of $\mathsf{ZFC}$, which in turn is strictly stronger than the mere consistency of $\mathsf{ZFC}$, but it is strictly weaker than the existence of inaccessible cardinals.

Now work in $L$, and consider a countable elementary substructure $Y\preceq L_\alpha$. Note that $Y$ is well-founded and satisfies $V=L$. Its collapse is then $L_\beta$ for some $\beta$ that, by necessity, is countable in $L$.

(Note that this also addresses (2), by taking $X=\emptyset$.)

By the way, this highlights some of the difficulties a challenger of "$V=L$" must face. Even if there are no inaccessibles, the model $M$ we began with may perfectly well be constructible, and satisfy that there are supercompact cardinals, or whatever. About this, you may also want to look at this post by Joel.

Andres Caicedo has answered Questions 1 and 2 as stated, but, just in case you intended $\kappa>\omega_1^L$ in Question 1 (since that's obviously necessary for the proposed conclusion), let me add that this is also sufficient for an affirmative answer to Question 1. The point is that, if an ordinal $\alpha$ is countable in $L$ then a bijection between $\alpha$ and $\omega$ appears in the construction of $L$ well before stage $\omega_1^L$. This is essentially contained in Gödel's proof of GCH in $L$. It uses much less than the full strength of ZFC in $L_\kappa$; you just need enough set theory to talk sensibly about countability of ordinals.